📝 Unit 2: Derivatives — Unit Test

Power, Product, Quotient, Chain, Implicit
✅ Graded — Counts Toward 70% Term Mark
⏱️ 75 min  |  Total: /60 marks
K/U
/15
Thinking
/15
Comm.
/15
Applic.
/15
Part A: Knowledge & Understanding [15]
1 [2]
Differentiate \(f(x)=5x^4-3\sqrt{x}+\frac{2}{x}\).
2 [2]
Find \(\frac{dy}{dx}\) for \(y=(x^2+3x)(x^3-1)\) using the product rule.
3 [3]
If \(g(x)=\dfrac{2x-1}{x^2+4}\), find \(g'(x)\) and \(g'(0)\).
4 [2]
MC: \(\frac{d}{dx}(x^2+1)^5\) at \(x=1\) equals:
5 [3]
Differentiate \(y=\sqrt{3x^2+5}\) and simplify.
6 [3]
Use implicit differentiation to find \(\frac{dy}{dx}\) for \(x^3+y^3=9xy\).
Part B: Thinking [15]
7 [5]
Find the equation of the tangent line to \(y=\dfrac{x}{x^2+1}\) at \(x=2\). Give answer in slope-intercept form.
8 [5]
Determine all values of \(x\) where the tangent line to \(f(x)=x^3-3x^2-9x+5\) is horizontal.
9 [5]
Find \(\frac{dy}{dx}\) for \(y=\frac{(2x+1)^3}{(x-1)^2}\). Use logarithmic differentiation OR a combination of rules. Show your method.
Part C: Communication [15]
10 [4]
Explain in detail why the product rule \((fg)'=f'g+fg'\) is NOT \(f'g'\). Show using \(f=x^2,\ g=x^3\) that the wrong rule gives a different answer.
11 [4]
Describe a step-by-step procedure for differentiating \(y=\sqrt{x^2+1}\) using the chain rule. Identify the "outer" and "inner" functions.
12 [4]
A student claims: "Implicit differentiation gives a different answer than solving for \(y\) and then differentiating." Address whether this is ever true. Use \(x^2+y^2=25\) at \((3,4)\) as a verification.
13 [3]
Compare the quotient rule and using the product rule on \(f\cdot g^{-1}\). Show they give the same result for \(\frac{x}{x+1}\).
Part D: Application [15]
14 [5]
Position: \(s(t)=t^3-6t^2+9t\) (m, s) for \(t\ge 0\). Find: a) velocity \(v(t)\); b) acceleration \(a(t)\); c) when is the particle momentarily at rest?
15 [5]
A demand function is \(p(x)=\dfrac{500}{x+10}\) (price in $; \(x\) units). Marginal revenue \(R'(x)\) where \(R(x)=x\cdot p(x)\). Compute \(R'(40)\) and interpret in context.
16 [5]
A circle \(x^2+y^2=100\) and a point \((6,8)\) on it. a) Find the slope of the tangent at \((6,8)\). b) Write the tangent line equation. c) Verify using the geometric fact that the tangent is perpendicular to the radius.

Evaluation Rubric

LevelDescription%
4Thorough, insightful80–100
3Considerable (provincial standard)70–79
2Some effectiveness60–69
1Limited effectiveness50–59
RInsufficient<50
📕 Answer Key

1. \(f'=20x^3-\frac{3}{2\sqrt{x}}-\frac{2}{x^2}\).

2. \(y'=(2x+3)(x^3-1)+(x^2+3x)(3x^2)=5x^4+12x^3-2x-3\).

3. \(g'=\frac{2(x^2+4)-(2x-1)(2x)}{(x^2+4)^2}=\frac{-2x^2+2x+8}{(x^2+4)^2}\); \(g'(0)=8/16=1/2\).

4. \(5(x^2+1)^4(2x)\); at 1: \(5(16)(2)=160\) → (b).

5. \(y'=\frac{6x}{2\sqrt{3x^2+5}}=\frac{3x}{\sqrt{3x^2+5}}\).

6. \(3x^2+3y^2 y'=9y+9x y'\) → \(y'=\frac{9y-3x^2}{3y^2-9x}=\frac{3y-x^2}{y^2-3x}\).

7. \(y'=\frac{1-x^2}{(x^2+1)^2}\); at 2: \(-3/25\). Point \((2,2/5)\). Tangent: \(y=-\frac{3}{25}x+\frac{16}{25}\).

8. \(f'=3x^2-6x-9=3(x-3)(x+1)\) → \(x=3,\,x=-1\).

9. \(\ln y=3\ln(2x+1)-2\ln(x-1)\); \(y'/y=\frac{6}{2x+1}-\frac{2}{x-1}\); \(y'=\frac{(2x+1)^3}{(x-1)^2}\!\left[\frac{6}{2x+1}-\frac{2}{x-1}\right]\).

10. \(f'g'=2x\cdot 3x^2=6x^3\), but \((x^5)'=5x^4\). Correct: \(f'g+fg'=2x\cdot x^3+x^2\cdot 3x^2=2x^4+3x^4=5x^4\). ✓

11. Outer = \(\sqrt{u}\), inner = \(u=x^2+1\). \(\frac{1}{2\sqrt{u}}\cdot u'=\frac{2x}{2\sqrt{x^2+1}}=\frac{x}{\sqrt{x^2+1}}\).

12. Same answer when both methods are valid. Implicit: \(y'=-x/y\); at (3,4): \(-3/4\). Solve: \(y=\sqrt{25-x^2}\), \(y'=-x/\sqrt{25-x^2}\); at 3: \(-3/4\). ✓

13. Quotient: \(\frac{(x+1)-x}{(x+1)^2}=\frac{1}{(x+1)^2}\). Product on \(x(x+1)^{-1}\): \(1\cdot(x+1)^{-1}+x\cdot(-1)(x+1)^{-2}=\frac{(x+1)-x}{(x+1)^2}=\frac{1}{(x+1)^2}\). ✓

14. a) \(v=3t^2-12t+9\); b) \(a=6t-12\); c) \(v=0\): \(3(t-1)(t-3)=0\) → \(t=1,3\) s.

15. \(R(x)=\frac{500x}{x+10}\); \(R'=\frac{500(x+10)-500x}{(x+10)^2}=\frac{5000}{(x+10)^2}\). \(R'(40)=5000/2500=2\) → revenue increases by ~$2 per additional unit at \(x=40\).

16. a) \(2x+2y y'=0\); at (6,8): \(y'=-6/8=-3/4\). b) \(y-8=-\frac{3}{4}(x-6)\). c) Radius slope = \(8/6=4/3\); product \(-3/4\cdot 4/3=-1\) → perpendicular ✓.